# Solution to Maxwell's equations in free space that are not plane waves [closed]

Are there solution to Maxwell's equations in free space that are not plane waves? I think there aren't. (Save trivial ones, i.e. E=const , B=const ) But i am not able to prove it. Please help. I would be grateful. :)

• What do you mean by "not plane waves"? Any superposition of solutions is again a solution. Are superpositions of plane waves (e.g. wavepackets) "plane waves" for you? – ACuriousMind Mar 9 '19 at 14:02
• @ACuriousMind Superposition of plane waves is another plane wave only? I think so.. Plane wave in the sense that the wavefront is a plane. – Jeevesh Juneja Mar 9 '19 at 14:05
• What about static homogenous fields? $\mathbf{E} = \text{const}, \mathbf{B} = \text{const}$. – Thomas Fritsch Mar 9 '19 at 14:09
• @ThomasFritsch Those are trivial. Any other? – Jeevesh Juneja Mar 9 '19 at 14:09
• see section on spherical waves in en.wikipedia.org/wiki/Wave_equation – hyportnex Mar 9 '19 at 14:44

As indicated, a combination of plane waves with different directions is not a plane wave. So, examples are countless (waves guides....)

You can take an electric field $$\underline{\overrightarrow{E}}=\underline{\overrightarrow{{{E}_{0}}}}{{e}^{j(\omega t-\overrightarrow{k}\cdot \overrightarrow{OM})}}$$ with a complex wave vector $$\overrightarrow{k}={{k}_{x}}\overrightarrow{{{e}_{x}}}-j{{k}_{y}}\overrightarrow{{{e}_{y}}}$$

In empty space, $${{\overrightarrow{k}}^{2}}={{k}_{x}}^{2}-{{k}_{y}}^{2}={{\omega }^{2}}/{{c}^{2}}$$ , and you'll have a nice non plane wave $$\overrightarrow{E}=\overrightarrow{{{E}_{0}}}{{e}^{-{{k}_{y}}y}}{{e}^{j(\omega t-{{k}_{x}}x)}}$$.

The magnetic field is simply $$\underline{\overrightarrow{B}}=\frac{1}{\omega }\overrightarrow{k}\wedge \underline{\overrightarrow{E}}$$

This wave arises naturally in the problem of total internal reflection. https://en.wikipedia.org/wiki/Total_internal_reflection

Sorry for my poor english.

The Maxwell equation in empty space is the wave equation. Plane waves form a basis for the solutions to these equations. However, it is also linear and you can build any solution that you can build from a superposition of these waves.

As to your specific question, if you want to proof this, you should grab a basic book on differential equations. The uniqueness of the basis for the solution space is ensured (plane waves) and the appropriate combination of them is fixed once provided with boundary conditions.

• The maxwell equation in empty space is not the wave equation. – Jeevesh Juneja Mar 9 '19 at 14:39
• – Jeevesh Juneja Mar 9 '19 at 14:41
• The Maxwell equations can be reduced the wave equations for the potential, that are homogeneous in the absence of charge and current. See physics.stackexchange.com/q/465013 – my2cts Mar 9 '19 at 15:21

There are solutions that are spherical or cylindrical waves:

In such cases, the amplitude of $$\vec E$$ goes like $$1/r$$ and $$1/\sqrt{r}$$ so the energy is conserved when integrated on the surface. The cylindrical waves are produced by a plane wave exiting a slit.

To be explicit, the spherical wave is a solution to the wave equation $$\nabla^2\psi( r,t)=\frac{1}{v^2}\frac{\partial ^2\psi( r,t)}{\partial t^2}$$ using spherical coordinates with $$\frac{\partial^2}{\partial r^2}(r\psi)=\frac{1}{v^2}\frac{\partial ^2 }{\partial t^2}(r\psi( r,t))\quad\Rightarrow\quad \psi(r,t)=\frac{f(r-vt)}{r}$$ and the cylindrical waves follows from solving in cylindrical coordinates.